Saturday, 15 February 2014

real analysis - Limit of Lp norm



Could someone help me prove that given a finite measure space (X,M,σ) and a measurable function f:XR in L and some Lq, limpfp=f?




I don't know where to start.


Answer



Fix δ>0 and let S_\delta:=\{x,|f(x)|\geqslant \lVert f\rVert_\infty-\delta\} for \delta<\lVert f\rVert_\infty. We have
\lVert f\rVert_p\geqslant \left(\int_{S_\delta}(\lVert f\rVert_\infty-\delta)^pd\mu\right)^{1/p}=(\lVert f\rVert_\infty-\delta)\mu(S_\delta)^{1/p},
since \mu(S_\delta) is finite and positive.
This gives
\liminf_{p\to +\infty}\lVert f\rVert_p\geqslant\lVert f\rVert_\infty.
As |f(x)|\leqslant\lVert f\rVert_\infty for almost every x, we have for p>q, \lVert f\rVert_p\leqslant\left(\int_X|f(x)|^{p-q}|f(x)|^qd\mu\right)^{1/p}\leqslant \lVert f\rVert_\infty^{\frac{p-q}p}\lVert f\rVert_q^{q/p},

giving the reverse inequality.


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